{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 标准的斐波那契数列求值过程"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def FN(a,b,n,mod):\n",
    "    if n == 1 or n == 2:\n",
    "        return 1\n",
    "    return (a * FN(a,b,n-1,mod) + b * FN(a,b,n-2,mod))%mod\n",
    "\n",
    "print(FN(1,1,10,10000))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 快速幂推导过程"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#数字快速幂 计算过程\n",
    "def quickpow(a,b):\n",
    "    d =1\n",
    "    if (b == 0): \n",
    "        return 1\n",
    "    while (b > 0):\n",
    "        if (b&0x01 == 1):\n",
    "            d = d * a\n",
    "        a = a * a\n",
    "        b = b>>1\n",
    "    return d\n",
    "\n",
    "print(quickpow(2,100))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "import math\n",
    "import numpy as np\n",
    "\n",
    "\n",
    "# 这个代码现在只能处理方阵\n",
    "def matrix22_quickpow(a,b):\n",
    "    d = np.eye(a.shape[0])\n",
    "    if (b == 0):\n",
    "        print(\"怎么会有0次方的矩阵?\")\n",
    "        return d\n",
    "    \n",
    "    while (b > 0):\n",
    "        if (b&0x01 == 1):\n",
    "            d = np.dot(d,a)\n",
    "        a = np.dot(a,a)\n",
    "        b = b>>1\n",
    "    return d\n",
    "\n",
    "\n",
    "def FNmatrix(a,b,n):\n",
    "    if ( n > 0 and  n <= 2):\n",
    "        return 1\n",
    "    npa = np.array([[a, b], [1,0]])\n",
    "    matrix_n1 = matrix22_quickpow(npa,n-1)\n",
    "    return int((matrix_n1[0][0]))\n",
    "\n",
    "\n",
    "def matrix22_quickpow_mod(a,b,modnum):\n",
    "    d = np.eye(a.shape[0])\n",
    "    if (b == 0):\n",
    "        #print(\"怎么会有0次方的矩阵?\")\n",
    "        return d\n",
    "    \n",
    "    while (b > 0):\n",
    "        if (b&0x01 == 1):\n",
    "            d = np.mod(np.dot(d,a),modnum)\n",
    "        a = np.mod(np.dot(a,a),modnum)\n",
    "        b = b>>1\n",
    "    return d\n",
    "\n",
    "\n",
    "def FNmod(a,b,n,modnum):\n",
    "    npa = np.array([[a, b], [1,0]])\n",
    "    matrix_n1 = matrix22_quickpow_mod(npa,n-1,modnum)\n",
    "    return int((matrix_n1[0][0]))\n",
    "\n",
    "\n",
    "for i in range(1,200):\n",
    "    print (FNmod(1,1,i,10), ' ', end='')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Largest prime factor\n",
    "\n",
    "The prime factors of 13195 are 5, 7, 13 and 29.<br>\n",
    "What is the largest prime factor of the number 600851475143 ?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [],
   "source": [
    "import math\n",
    "\n",
    "# 朴素求质数方式\n",
    "def FUNC_judgePrime_simple(num):\n",
    "    if num == 2:\n",
    "        return True\n",
    "    if num == 1:\n",
    "        return False\n",
    "    for i in range(1,num+1):\n",
    "        if num % i == 0 and ( i != 1 and i != num ):\n",
    "            # print('first factor is ', i)\n",
    "            return False\n",
    "    return True\n",
    "\n",
    "\n",
    "# 一个数字在进行因式分解时(只有两个数字的情况下，其他也符合，只拿这个作为比较)，一侧必小于等于sqrt(num)，另一侧必大于等于sqrt(num)\n",
    "def FUNC_judgePrime_sqrt(num):\n",
    "    if num == 1:\n",
    "        return False\n",
    "    if (num == 2 or num == 3):\n",
    "        return True\n",
    "    temp = int(math.sqrt(num))\n",
    "    for i in range(2, temp+1):\n",
    "        if num % i == 0:\n",
    "            return False\n",
    "    return True"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "sum is  142913828922\n"
     ]
    }
   ],
   "source": [
    "# num=127\n",
    "# print('num ', num, ' Prime is ',FUNC_judgePrime_simple(num))\n",
    "# print('num ', num, ' Prime is ',FUNC_judgePrime_sqrt(num))\n",
    "\n",
    "def FUNC_getPrime_range(num):\n",
    "    for i in range(2,num+1):\n",
    "        # if FUNC_judgePrime_simple(i) is True:\n",
    "        if FUNC_judgePrime_sqrt(i) is True:\n",
    "            #print(\"get prime \", i)\n",
    "            print(i,' ', end='')\n",
    "\n",
    "#FUNC_getPrime_range(10000)\n",
    "\n",
    "def FUNC_getSum_of_Primes_of_twoMillion(num):\n",
    "    sum = 0\n",
    "    for i in range(2,num+1):\n",
    "        if FUNC_judgePrime_sqrt(i) is True:\n",
    "            sum += i\n",
    "    print(\"sum is \",sum)\n",
    "    \n",
    "FUNC_getSum_of_Primes_of_twoMillion(2000000)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "Prime_100=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]\n",
    "Prime_10000=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257,3259,3271,3299,3301,3307,3313,3319,3323,3329,3331,3343,3347,3359,3361,3371,3373,3389,3391,3407,3413,3433,3449,3457,3461,3463,3467,3469,3491,3499,3511,3517,3527,3529,3533,3539,3541,3547,3557,3559,3571,3581,3583,3593,3607,3613,3617,3623,3631,3637,3643,3659,3671,3673,3677,3691,3697,3701,3709,3719,3727,3733,3739,3761,3767,3769,3779,3793,3797,3803,3821,3823,3833,3847,3851,3853,3863,3877,3881,3889,3907,3911,3917,3919,3923,3929,3931,3943,3947,3967,3989,4001,4003,4007,4013,4019,4021,4027,4049,4051,4057,4073,4079,4091,4093,4099,4111,4127,4129,4133,4139,4153,4157,4159,4177,4201,4211,4217,4219,4229,4231,4241,4243,4253,4259,4261,4271,4273,4283,4289,4297,4327,4337,4339,4349,4357,4363,4373,4391,4397,4409,4421,4423,4441,4447,4451,4457,4463,4481,4483,4493,4507,4513,4517,4519,4523,4547,4549,4561,4567,4583,4591,4597,4603,4621,4637,4639,4643,4649,4651,4657,4663,4673,4679,4691,4703,4721,4723,4729,4733,4751,4759,4783,4787,4789,4793,4799,4801,4813,4817,4831,4861,4871,4877,4889,4903,4909,4919,4931,4933,4937,4943,4951,4957,4967,4969,4973,4987,4993,4999,5003,5009,5011,5021,5023,5039,5051,5059,5077,5081,5087,5099,5101,5107,5113,5119,5147,5153,5167,5171,5179,5189,5197,5209,5227,5231,5233,5237,5261,5273,5279,5281,5297,5303,5309,5323,5333,5347,5351,5381,5387,5393,5399,5407,5413,5417,5419,5431,5437,5441,5443,5449,5471,5477,5479,5483,5501,5503,5507,5519,5521,5527,5531,5557,5563,5569,5573,5581,5591,5623,5639,5641,5647,5651,5653,5657,5659,5669,5683,5689,5693,5701,5711,5717,5737,5741,5743,5749,5779,5783,5791,5801,5807,5813,5821,5827,5839,5843,5849,5851,5857,5861,5867,5869,5879,5881,5897,5903,5923,5927,5939,5953,5981,5987,6007,6011,6029,6037,6043,6047,6053,6067,6073,6079,6089,6091,6101,6113,6121,6131,6133,6143,6151,6163,6173,6197,6199,6203,6211,6217,6221,6229,6247,6257,6263,6269,6271,6277,6287,6299,6301,6311,6317,6323,6329,6337,6343,6353,6359,6361,6367,6373,6379,6389,6397,6421,6427,6449,6451,6469,6473,6481,6491,6521,6529,6547,6551,6553,6563,6569,6571,6577,6581,6599,6607,6619,6637,6653,6659,6661,6673,6679,6689,6691,6701,6703,6709,6719,6733,6737,6761,6763,6779,6781,6791,6793,6803,6823,6827,6829,6833,6841,6857,6863,6869,6871,6883,6899,6907,6911,6917,6947,6949,6959,6961,6967,6971,6977,6983,6991,6997,7001,7013,7019,7027,7039,7043,7057,7069,7079,7103,7109,7121,7127,7129,7151,7159,7177,7187,7193,7207,7211,7213,7219,7229,7237,7243,7247,7253,7283,7297,7307,7309,7321,7331,7333,7349,7351,7369,7393,7411,7417,7433,7451,7457,7459,7477,7481,7487,7489,7499,7507,7517,7523,7529,7537,7541,7547,7549,7559,7561,7573,7577,7583,7589,7591,7603,7607,7621,7639,7643,7649,7669,7673,7681,7687,7691,7699,7703,7717,7723,7727,7741,7753,7757,7759,7789,7793,7817,7823,7829,7841,7853,7867,7873,7877,7879,7883,7901,7907,7919,7927,7933,7937,7949,7951,7963,7993,8009,8011,8017,8039,8053,8059,8069,8081,8087,8089,8093,8101,8111,8117,8123,8147,8161,8167,8171,8179,8191,8209,8219,8221,8231,8233,8237,8243,8263,8269,8273,8287,8291,8293,8297,8311,8317,8329,8353,8363,8369,8377,8387,8389,8419,8423,8429,8431,8443,8447,8461,8467,8501,8513,8521,8527,8537,8539,8543,8563,8573,8581,8597,8599,8609,8623,8627,8629,8641,8647,8663,8669,8677,8681,8689,8693,8699,8707,8713,8719,8731,8737,8741,8747,8753,8761,8779,8783,8803,8807,8819,8821,8831,8837,8839,8849,8861,8863,8867,8887,8893,8923,8929,8933,8941,8951,8963,8969,8971,8999,9001,9007,9011,9013,9029,9041,9043,9049,9059,9067,9091,9103,9109,9127,9133,9137,9151,9157,9161,9173,9181,9187,9199,9203,9209,9221,9227,9239,9241,9257,9277,9281,9283,9293,9311,9319,9323,9337,9341,9343,9349,9371,9377,9391,9397,9403,9413,9419,9421,9431,9433,9437,9439,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887,9901,9907,9923,9929,9931,9941,9949,9967,9973]\n",
    "\n",
    "# 取巧了\n",
    "def FUNC_primefactoring(num):\n",
    "    new_num = num\n",
    "    while new_num > 1:\n",
    "        #for i in Prime_100:\n",
    "        for i in Prime_10000:\n",
    "            if new_num % i == 0:\n",
    "                new_num = new_num / i\n",
    "                print(i, ' ', end='')\n",
    "                break\n",
    "\n",
    "FUNC_primefactoring(600851475143)            "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "scrolled": true
   },
   "outputs": [],
   "source": [
    "# 钱币分类, 假设有1元、2元、5元、10元、20元、50元、100的纸币分别为c0, c1, c2, c3, c4, c5, c6,张。\n",
    "# 现在要用这些钱来支付K元，至少要用多少张纸币？如果能找，则输出纸币的张数，不能找则输出No\n",
    "\n",
    "money_list=  [100, 50, 20, 10, 5, 2, 1]\n",
    "money_limit= [5,   0,  1,  2,  5, 5, 5]\n",
    "money_num=   [0,   0,  0,  0,  0, 0, 0]\n",
    "\n",
    "\n",
    "# 当前算法是没有限制每个纸币的数目的\n",
    "def FUNC_money_result(money):\n",
    "    new_money=money\n",
    "    money_len = len(money_list)\n",
    "    for i in range(money_len):\n",
    "        num = int(new_money/money_list[i])\n",
    "        new_money = new_money % money_list[i]\n",
    "        if num == 0:\n",
    "            continue\n",
    "        else:\n",
    "            money_num[i] = num\n",
    "    numsum = 0\n",
    "    for i in range(money_len):\n",
    "         numsum += money_num[i]\n",
    "    if numsum == 0:\n",
    "        print(\"NO\")\n",
    "    else:\n",
    "        for i in range(money_len):\n",
    "            if money_num[i] != 0:\n",
    "                print (' ',money_list[i],' is ', money_num[i],end=\"\")\n",
    "    \n",
    "# 加上纸币数目限制\n",
    "def FUNC_money_limit_result(money):\n",
    "    new_money=money\n",
    "    money_len = len(money_list)\n",
    "    for i in range(money_len):\n",
    "        num = int(new_money/money_list[i])\n",
    "        if num > money_limit[i]:\n",
    "            num = money_limit[i]\n",
    "        new_money = new_money - money_list[i] * num\n",
    "        if num == 0:\n",
    "            continue\n",
    "        else:\n",
    "            money_num[i] = num\n",
    "        #最后用到1元了，但是还是有钱没用完，那说明这个方案无法继续了。\n",
    "        if i == money_len-1 and new_money != 0: \n",
    "            print(\"NO\")\n",
    "            return\n",
    "    numsum = 0\n",
    "    for i in range(money_len):\n",
    "         numsum += money_num[i]\n",
    "    if numsum == 0:\n",
    "        print(\"NO\")\n",
    "    else:\n",
    "        for i in range(money_len):\n",
    "            if money_num[i] != 0:\n",
    "                print (money_list[i],' * ', money_num[i],'+',end=\"\")    \n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#FUNC_money_result(13)\n",
    "FUNC_money_limit_result(59)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 数字回文\n",
    "\n",
    "求 两个三位数构造的最大回文数字是哪个，三位数从100-999，最小到最大是 100 \\* 100 ~~ 999 \\* 999<br>\n",
    "怎么判断一个数字是否是回文。先把数字字符串化，然后求得字符串中间位置，str\\[0\\] ~ str\\[mid\\] 和 str\\[mid+1\\] ~ str\\[end\\]，需要注意长度，奇数和偶数"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# 判断一个字符串是否是回文\n",
    "def FUNC_palindrome_string(string):\n",
    "    slen=len(string)\n",
    "    mid = int(slen/2)\n",
    "    for i in range(0,mid): #实际到mid-1\n",
    "        if string[i] != string[slen-1-i]:\n",
    "            return False\n",
    "    return True\n",
    "\n",
    "# 将数字转换成字符串 python方式\n",
    "def FUNC_num_to_string(number):\n",
    "    return str(number)\n",
    "\n",
    "def FUNC_num_get_AxB(number,nmin,nmax):\n",
    "    for i in range(nmax,nmin-1,-1):\n",
    "        ri = int(number/i)\n",
    "        if number % i == 0 and ri >= nmin and ri <= nmax:\n",
    "            print(\"found result is \", number, i, ri)\n",
    "            return True\n",
    "        else:\n",
    "            continue\n",
    "    return False"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# 在给定数字范围内找到一个最大的回文数字 \n",
    "def FUNC_find_max_palindrome_number(min,max,nmin,nmax):\n",
    "    for i in range(max,min-1,-1):\n",
    "        # 找到的数字有可能是质数，无法进行因式分解，也有可能数字本身不能成为100*100这种结果\n",
    "        if FUNC_palindrome_string(FUNC_num_to_string(i)) is True: \n",
    "            # print('found max num is ',i)\n",
    "            if FUNC_num_get_AxB(i,nmin,nmax) is True:\n",
    "                break"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "import time\n",
    "# 测试字符串回文是否正确\n",
    "# stringlist=['1231321','12344321','1112345678']\n",
    "# for string in stringlist:\n",
    "#     print('string ',string, ' palindrome ', FUNC_palindrome_string(string))\n",
    "start = time.clock()  \n",
    "FUNC_find_max_palindrome_number(100*100,999*999,100,999)\n",
    "end = time.clock()\n",
    "print(\"The function run time is : %.03f seconds\" %(end-start))\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "result is  25164150\n"
     ]
    }
   ],
   "source": [
    "# Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.\n",
    "suma=5050*5050\n",
    "sumb=0\n",
    "for i in range(1,101):\n",
    "    sumb += i*i\n",
    "print(\"result is \",suma-sumb)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "23514624000\n"
     ]
    }
   ],
   "source": [
    "str=\"73167176531330624919225119674426574742355349194934\\\n",
    "96983520312774506326239578318016984801869478851843\\\n",
    "85861560789112949495459501737958331952853208805511\\\n",
    "12540698747158523863050715693290963295227443043557\\\n",
    "66896648950445244523161731856403098711121722383113\\\n",
    "62229893423380308135336276614282806444486645238749\\\n",
    "30358907296290491560440772390713810515859307960866\\\n",
    "70172427121883998797908792274921901699720888093776\\\n",
    "65727333001053367881220235421809751254540594752243\\\n",
    "52584907711670556013604839586446706324415722155397\\\n",
    "53697817977846174064955149290862569321978468622482\\\n",
    "83972241375657056057490261407972968652414535100474\\\n",
    "82166370484403199890008895243450658541227588666881\\\n",
    "16427171479924442928230863465674813919123162824586\\\n",
    "17866458359124566529476545682848912883142607690042\\\n",
    "24219022671055626321111109370544217506941658960408\\\n",
    "07198403850962455444362981230987879927244284909188\\\n",
    "84580156166097919133875499200524063689912560717606\\\n",
    "05886116467109405077541002256983155200055935729725\\\n",
    "71636269561882670428252483600823257530420752963450\"\n",
    "\n",
    "max = 0\n",
    "cur = 0\n",
    "numlen=13\n",
    "for i in range(1,len(str)-numlen):\n",
    "    cur=1\n",
    "    for j in range(0,numlen):\n",
    "        cur *= int(str[i+j])\n",
    "        if cur > max:\n",
    "            max = cur\n",
    "print(max)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 图论-数据结构\n",
    "\n",
    "参考链接  https://zhuanlan.zhihu.com/p/25498681"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.6.4"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}
